Since it was deciphered four decades ago, some have claimed

Since it was deciphered four decades ago, some have claimed that the genetic code must be a frozen accident, while others have argued that it was shaped by natural selection. A striking feature of the genetic code is its inherent resistance to the effects of mutation. For example, a change in the third position of a codon often specifies the same amino acid or one with similar chemical properties.

The natural code resists mutation more effectively (is less susceptible to error) than most other possible versions, as illustrated in Figure Q1–1. Only one in a million computer generated “random” codes is more error-resistant than the natural genetic code. Does the extraordinary mutation resistance of the genetic code argue in favor of its origin as a frozen accident or as a result of natural selection? Explain your reasoning.

Figure Q1–1

When plant hemoglobin genes were first discovered in legumes, it

When plant hemoglobin genes were first discovered in legumes, it was so surprising to find a gene typical of animal blood that it was hypothesized that the plant gene arose by horizontal transfer from an animal. Many more hemoglobin genes have now been sequenced, and a phylogenetic tree based on some of these sequences is shown in Figure Q1–2.

A. Does this tree support or refute the hypothesis that the plant hemoglobins arose by horizontal gene transfer?

B. Supposing that the plant hemoglobin genes were originally derived from a parasitic nematode, for example, what would you expect the phylogenetic tree to look like?

Figure Q1–1

Figure Q1–2

Rous sarcoma virus (RSV) carries an oncogene called Src, which

Rous sarcoma virus (RSV) carries an oncogene called Src, which encodes a continuously active protein tyrosine kinase that leads to unchecked cell proliferation. Normally, Src carries an attached fatty acid (myristoylate) group that allows it to bind to the cytoplasmic side of the plasma membrane. A mutant version of Src that does not allow attachment of myristoylate does not bind to the membrane. Infection of cells with RSV encoding either the normal or the mutant form of Src leads to the same high level of protein tyrosine kinase activity, but the mutant Src does not cause cell proliferation.

A. Assuming that the normal Src is all bound to the plasma membrane and that the mutant Src is distributed throughout the cytoplasm, calculate their relative concentrations in the neighborhood of the plasma membrane. For the purposes of this calculation, assume that the cell is a sphere with a radius (r) of 10 μm and that the mutant Src is distributed throughout the cell, whereas the normal Src is confined to a 4-nm-thick layer immediately beneath the membrane. [For this problem, assume that the membrane has no thickness. The volume of a sphere is (4/3)πr3.] 

B. The target (X) for phosphorylation by Src resides in the membrane. Explain why the mutant Src does not cause cell proliferation.

Titin, which has a molecular weight of about 3 ×

Titin, which has a molecular weight of about 3 × 106, is the largest polypeptide yet described. Titin molecules extend from muscle thick filaments to the Z disc; they are thought to act as springs to keep the thick filaments centered in the sarcomere. Titin is composed of a large number of repeated immunoglobulin (Ig) sequences of 89 amino acids, each of which is folded into a domain about 4 nm in length (Figure Q3–2A).

You suspect that the springlike behavior of titin is caused by the sequential unfolding (and refolding) of individual Ig domains. You test this hypothesis using the atomic force microscope, which allows you to pick up one end of a protein molecule and pull with an accurately measured force. For a fragment of titin containing seven repeats of the Ig domain, this experiment gives the sawtooth force-versus- extension curve shown in Figure Q3–2B. If the experiment is repeated in a solution of 8 M urea (a protein denaturant), the peaks disappear and the measured extension becomes much longer for a given force. If the experiment is repeated after the protein has been cross-linked by treatment with glutaraldehyde, once again the peaks disappear but the extension becomes much smaller for a given force.

Figure Q3–2A

A. Are the data consistent with your hypothesis that titin’s springlike behavior is due to the sequential unfolding of individual Ig domains? Explain your reasoning.

B. Is the extension for each putative domain-unfolding event the magnitude you would expect? (In an extended polypeptide chain, amino acids are spaced at intervals of 0.34 nm.)

C. Why is each successive peak in Figure Q3–2B a little higher than the one before?

D. Why does the force collapse so abruptly after each peak?

Consider the following statement. “To produce one molecule of each

Consider the following statement. “To produce one molecule of each possible kind of polypeptide chain, 300 amino acids in length, would require more atoms than exist in the universe.” Given the size of the universe, do you suppose this statement could possibly be correct? Since counting atoms is a tricky business, consider the problem from the standpoint of mass. The mass of the observable universe is estimated to be about 1080 grams, give or take an order of magnitude or so. Assuming that the average mass of an amino acid is 110 daltons, what would be the mass of one molecule of each possible kind of polypeptide chain 300 amino acids in length? Is this greater than the mass of the universe?

“Diffusion” sounds slow—and over everyday distances it is—but on the

“Diffusion” sounds slow—and over everyday distances it is—but on the scale of a cell it is very fast. The average instantaneous velocity of a particle in solution—that is, the velocity between the very frequent collisions—is

v = (kT/m)½

where k = 1.38 × 10–16 g cm2/K sec2, T = temperature in K (37°C is 310 K), and m = mass in g/molecule.

Calculate the instantaneous velocity of a water molecule (molecular mass = 18 daltons), a glucose molecule (molecular mass = 180 daltons), and a myoglobin molecule (molecular mass = 15,000 daltons) at 37°C. Just for fun, convert these numbers into kilometers/hour.

Before you do any calculations, try to guess whether the molecules are moving at a slow crawl (<1 km/hr), an easy walk (5 km/hr), or a record-setting sprint (40 km/hr).

Does a Snickers™ candy bar (65 g, 1360 kJ) provide

Does a Snickers™ candy bar (65 g, 1360 kJ) provide enough energy to climb from Zermatt (elevation 1660 m) to the top of the Matterhorn (4478 m, Figure Q2–3), or might you need to stop at HÖrnli Hut (3260 m) to eat another one? Imagine that you and your gear have a mass of 75 kg, and that all of your work is done against gravity (that is, you are just climbing straight up). Remember from your introductory physics course that

work (J) = mass (kg) × g (m/sec2) × height gained (m)

where g is acceleration due to gravity (9.8 m/sec2). One joule is 1 kg m2/sec2.

What assumptions made here will greatly underestimate how much candy you need?

Figure Q2–3

 

The molecular weight of ethanol (CH3CH2OH) is 46 and its

The molecular weight of ethanol (CH3CH2OH) is 46 and its density is 0.789 g/cm3.

A. What is the molarity of ethanol in beer that is 5% ethanol by volume? [Alcohol content of beer varies from about 4% (lite beer) to 8% (stout beer).]

B. The legal limit for a driver’s blood alcohol content varies, but 80 mg of ethanol per 100 mL of blood (usually referred to as a blood alcohol level of 0.08) is typical. What is the molarity of ethanol in a person at this legal limit?

C. How many 12-oz (355-mL) bottles of 5% beer could a 70-kg person drink and remain under the legal limit? A 70-kg person contains about 40 liters of water. Ignore the metabolism of ethanol, and assume that the water content of the person remains constant.

D. Ethanol is metabolized at a constant rate of about 120 mg per hour per kg body weight, regardless of its concentration. If a 70-kg person were at twice the legal limit (160 mg/100 mL), how long would it take for their blood alcohol level to fall below the legal limit?

The protein SmpB binds to a special species of tRNA,

The protein SmpB binds to a special species of tRNA, tmRNA, to eliminate the incomplete proteins made from truncated mRNAs in bacteria. If the binding of SmpB to tmRNA is plotted as fraction tmRNA bound versus SmpB concentration, one obtains a symmetrical S-shaped curve as shown in Figure Q3–3. This curve is a visual display of a very useful relationship between Kand concentration, which has broad applicability. The general expression for fraction of ligand bound is derived from the equation for Kd (Kd = [Pr][L]/[Pr–L]) by substituting ([L]TOT – [L]) for [Pr–L] and rearranging. Because the total concentration of ligand ([L]TOT) is equal to the free ligand ([L]) plus bound ligand ([Pr–L]), fraction bound = [Pr–L]/[L]TOT = [Pr]/([Pr] + Kd)

Figure Q3-3

For SmpB and tmRNA, the fraction bound = [SmpB– tmRNA]/[tmRNA]TOT = [SmpB]/([SmpB] + Kd). Using this relationship, calculate the fraction of tmRNA bound for SmpB concentrations equal to 104 Kd, 103 Kd, 102 Kd, 101 Kd, Kd, 10–1 Kd, 10–2 Kd, 10–3 Kd, and 10–4 Kd.

Look at the two yeast colonies in Figure Q4–3. Each

Look at the two yeast colonies in Figure Q4–3. Each of these colonies contains about 100,000 cells descended from a single yeast cell, originally somewhere in the middle of the clump. A white colony arises when the Ade2 gene is expressed from its normal chromosomal location. When the Ade2 gene is moved to a location near a telomere, it is packed into heterochromatin and inactivated in most cells, giving rise to colonies that are mostly red. In these largely red colonies, white sectors fan out from the middle of the colony. In both the red and white sectors, the Ade2 gene is still located near telomeres. Explain why white sectors have formed near the rim of the red colony. Based on the patterns observed, what can you conclude about the propagation of the transcriptional state of the Ade2 gene from mother to daughter cells in this experiment?

Figure Q4–3